Answer:
Letter A. [tex]y=4.55 m[/tex]
Explanation:
Let's use the wave equation:
[tex]y=Asin(kx-\omega t)[/tex]
A is the amplitude (A=6.44 m)t is the time (t=0.71 s)k is the wave number (k=2.34 1/m)ω is the angular frequency (ω=2.88 rad/s)x is the propagation of the x direction (x=1.21 m)Therefore the displacement y will be:
[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]
[tex]y=4.55 m[/tex]
The answer is letter A.
I hope it helps you!
Answer:
Explanation:
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
Amplitude (A) of the simple harmonic wave = 6.44 m
wave number (k) of the given wave = 2.34 m-1
Angular frequency (ω) of the given wave = 2.88 rad/s
Displacement x = 1.21 m and time t = 0.71 s
Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by
y = Asin(kx - ωt)
= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]
Y=6.44sin(0.7866 rad)
0.7866rad*(180 degrees/pi rad) =45.1
Y=6.44sin(45.1)
Y=4.55m
Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?
Explanation:
The relation between resistance and resistivity is given by :
[tex]R=\rho \dfrac{l}{A}[/tex]
[tex]\rho[/tex] is resistivity of material
l is length of wire
A is area of cross section of wire
Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is
Why do some nucleus emit electrons?
Answer:
In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Explanation:
Hope this helps!
John pushes Hector on a plastic toboggan.The free-body diagram is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = negative 490 N. The second vector is pointing right, labeled F Subscript t Baseline = 735 N. The third vector is pointing upward, labeled F Subscript N Baseline = 490 N. The fourth vector is pointing left, labeled F Subscript f Baseline = negative 245 N. The up and down vectors are the same length. The right vector is longer than the left vector. What is the net force acting on Hector and the toboggan?
Answer:
490 N
Explanation:
is the correct answer
If the up and down vectors are the same length. The right vector is longer than the left vector, then the net force acting on Hector and the toboggan would be 490 Newtons.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
As given in the problem John pushes Hector on a plastic toboggan .The free-body diagram is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = negative 490 N. The second vector is pointing right, labeled F Subscript t Baseline = 735 N. The third vector is pointing upward, labeled F Subscript N Baseline = 490 N. The fourth vector is pointing left, labeled F Subscript f Baseline = negative 245 N.
The net force acting on the vertical direction = 490-490
=0
The net force acting on the horizontal direction = 735 -245
=490 Newtons
Thus, the net force acting on Hector and the toboggan would be 490 Newtons.
Learn more about Newton's second law from here, refer to the link ;
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A single loop of wire with an area of 0.0820 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.220 T/s .
Required:
a. What emf is induced in this loop?
b. If the loop has a resistance of 0.600Ω, find the current induced in the loop.
Answer:
a) emf = 0.01804 V
b) I = 0.03 A
Explanation:
a) The emf is calculated by using the following formula:
[tex]|emf|=|\frac{d\Phi_B}{dt}|=|\frac{d(A\cdot B)}{dt}|[/tex] [tex]=A|\frac{dB}{dt}|[/tex]
A: area of the loop = 0.0820m^2
B: magnitude of the magnetic field
dB/dt: change of the magnetic field, in time: 0.220 T/s
Where ФB is the magnetic flux, the surface vector and magnetic vector are perpendicular between them, and the area A is constant.
You replace the values of A and dB/dt in the equation (1):
[tex]|emf|=(0.082m^2)(0.220T/s)=0.01804V[/tex]
b) The current in the loop is:
[tex]I=\frac{emf}{R}[/tex]
R: resistance of the loop = 0.600Ω
[tex]I=\frac{0.01804V}{0.600\Omega}=0.03A=30mA[/tex]
a. The emf induced in this loop is 18.04mV.
b. The current induced in the loop is 30.06mA.
a. We know that,
[tex]flux(\phi)=B*A[/tex]
Where B is magnetic field and A is the area.
[tex]emf=\frac{d\phi}{dt}=A*\frac{dB}{dt}[/tex]
Given that, Area , [tex]A=0.0820m^{2},B=3.80T,\frac{dB}{dt}=0.220T/s[/tex]
Substituting all values in above equation.
[tex]emf=0.0820*0.220=0.01804V=18.04mV[/tex]
b. Resistance, [tex]R=0.600ohm[/tex]
Current induced in the loop is,
[tex]I=\frac{emf}{R}=18.04/0.6=30.06mA[/tex]
Hence, the emf induced in this loop is 18.04mV.
The current induced in the loop is 30.06mA.
Learn more:
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Which of the following statements is true of a gas?
It has a fixed volume, but not a fixed shape
It has closely packed molecules
It can change into a liquid by adding heat
It takes the shape and size of a container
Answer:
it takes the shape and size of the container that it is in
Explanation:
Answer:
it takes the shape and size of a container
please help! i will be giving 50 points, this is for my psychology class.
Iris has been ahead of her classmates for as long as she has been in school. Lately, her classmates have started making fun of her for being a “teacher’s pet,” and they mock her whenever she raises her hand to answer a question.
Iris is most likely being negatively stereotyped as being __________.
A.
below average
B.
normal
C.
intellectually disabled
D.
gifted
Answer:
D
Explanation:
the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)
Answer:
D
Explanation:
state Ohm`s law as applied in electricity
Answer:
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.
The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichit is made has an index of refraction of 1.38, and its front surface is convex,with a radius of curvature of 5.00 {\rm mm}.(Note: The results obtained here are not strictlyaccurate, because, on one side, the cornea has a fluid with arefractive index different from that of air.)a) If this focal length is in air, what is the radius ofcurvature of the back side of the cornea? (in mm)b) The closest distance at which a typical person can focus onan object (called the near point) is about 25.0 {\rm cm}, although this varies considerably with age. Wherewould the cornea focus the image of an 10.0 {\rm mm}-tall object at the near point? (in mm)c) What is the height of the image in part B? (mm)d) Is this image real or virtual? Is it erect orinverted?
Answer:
Explanation:
a )
from lens makers formula
[tex]\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})[/tex]
f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face
putting the values
[tex]\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})[/tex]
1.462 = 2 - 1 / r₂
1 / r₂ = .538
r₂ = 1.86 cm .
= 18.6 mm .
b )
object distance u = 25 cm
focal length of convex lens f = 1.8 cm
image distance v = ?
lens formula
[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}[/tex]
[tex]\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}[/tex]
.5555 - .04
= .515
v = 1.94 cm
c )
magnification = v / u
= 1.94 / 25
= .0776
size of image = .0776 x size of object
= .0776 x 10 mm
= .776 mm
It will be a real image and it will be inverted.
Use the Lab screen to expand your ideas about what affects the landing location and path of a projectile. List any discoveries you made to identify additional things that affect the landing site of a projectile and/or path of a projectile. Next to each item, briefly explain why you think the motion of the projectile is affected..
Answer:
* air resistance.
*the direction of the rotation of the Earth
rotation of the thrown body
Explanation:
The projectile launch is described by the expressions
x-axis x = v₀ₓ t
y-axis y = [tex]v_{oy}[/tex] t - ½ gt²
When the things that affect this movement are analyzed, in order of importance we have:
* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.
* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.
* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption
* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height
An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
1.been both -ve charged or both +be charged particles
2. 3.52mC
Explanation:
For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.
Now the Force sustain by the extended string is
F = Ke;
Where K is the force constant of the string, 320 N/m
e is the extension,0.033 m
F = 320 × 0.033 =10.56N
2.But according to columns law of charge;
F = kQ1 Q2
But Q1=Q2{ since the charge are of the same magnitude}.
Hence F = KQ^2
Where K is columns constant =9×10^9F/m
Hence Q=√F/K
Q= √10.56/9×10^9
=3.52×10^-3C
= 3.52mC
A soccer player is benched for being late to the game. In a fit of anger, she drops her ball from the top of the Physics building. It falls 4.9 meters after 1.0 second has elapsed. How much farther does it fall in the next 2.0 seconds
Answer:
The distance is [tex]S = 39.2 \ m[/tex]
Explanation:
From the question we are told that
The distance covered after t = 1 s is [tex]d = 4.9 \ m[/tex]
According to the equation of motion
[tex]v^2 = u^2 + 2ad[/tex]
Now u = 0 m/s since before the drop the ball was at rest
[tex]v^2 = 2ad[/tex]
here [tex]a =g = 9.8 \ m/s^2[/tex]
So
[tex]v = 9.8 m/s[/tex]
Also from equation of motion we have that
[tex]S = ut + \frac{1}{2} at^2[/tex]
Now at t = 2 s , as given from the question
Then u = v = 9.8 m/s
And
[tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]
[tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]
[tex]S = 39.2 \ m[/tex]
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.
mQ: the mass of the quarterback
mB: the mass of the football
(vQx)i: the horizontal velocity of quarterback before throwing the ball
(vBx)i: the horizontal velocity of football before being thrown
(vQx)f: the horizontal velocity of quarterback after throwing the ball
(vBx)f: the horizontal velocity of football after being thrown
Answer:
vBxf = 0.08625m/s
Explanation:
This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.
[tex]p_f=p_i[/tex]
pf: final momentum
pi: initial momentum
The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.
Then, you have:
[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex] (1)
mQ: the mass of the quarterback = 80kg
mB: the mass of the football = 0.43kg
(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s
(vBx)i: the horizontal velocity of football before being thrown = 0m/s
(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?
(vBx)f: the horizontal velocity of football after being thrown = 15 m/s
You replace the values of the variables in the equation (1), and you solve for (vBx)f:
[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]
Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.
Hence, the speed of the quarterback after he throws the ball is 0.08625m/s
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which would allow astronauts to experience a sort of artificial gravity when walking along the inner wall of the station's outer rim. (a) Imagine one such station with a diameter of 104 m, where the apparent gravity is 2.20 m/s2 at the outer rim. How fast is the station rotating in revolutions per minute
Answer:
f = 1.96 revolutions per minute
Explanation:
The formula for the the frequency of revolution of a satellite, to develop an artificial gravity, with the help of centripetal acceleration is given as follows:
f = (1/2π)√(ac/r)
where,
f = frequency of rotation = ?
ac = centripetal acceleration= apparent gravity or artificial gravity = 2.2 m/s²
r = radius of station or satellite = diameter/2 = 104 m/2 = 52 m
Therefore,
f = (1/2π)√[(2.2 m/s²)/(52 m)]
f = (0.032 rev/s)(60 s/min)
f = 1.96 revolutions per minute
A river flows due south with a speed of 5.00 m/s. A man steers a motorboat across the river; his velocity relative to the water is 4.00 m/s due east. The river is 780 m wide. Part A What is the magnitude of his velocity relative to the earth
Answer:
6.4 m/s
Explanation:
From the question, we are given that
Speed of the river, v(r) = 5 m/s
velocity relative to the water, v(w) = 4 m/s
Width of the river, d = 780 m
The magnitude of his velocity relative to the earth is v(m)
v(m) can be gotten by using the relation
[v(m)]² = [v(w)]² + [v(r)]²
[v(m)]² = 4² + 5²
[v(m)]² = 16 + 25
[v(m)]² = 41
v(m) = √41
v(m) = 6.4 m/s
thus, the magnitude of the velocity relative to earth is 6.4 m/s
To move a large crate across a rough floor, you push on it with a force at an angle of 15 degrees below the horizontal. Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.49.
Answer:
663N
Explanation:
We need to find the force that will overcome the frictional force.
The angle of the normal force is 15°.
The mass of the crate is 32 kg
The coefficient of static friction is 0.49
Frictional force is given in terms of Normal force as:
F = μNcosθ
where μ = coefficient of static friction
N = normal force
θ = angle of normal force
Frictional force is given as:
F = mg
=>mg = μNcosθ
=> N = mg/(μcosθ)
N = (32 * 9.8) / (0.49 * cos15)
N= 313.6 / 0.473
N = 663 N
The force needed to cause the box to move must be 663N or greater.
an object's resistance to any change in motion is the_________ of the object.
An object's resistance to any change in motion is the Inertia of the object.
Convert from scientific notation to standard form
9.512 x 10-8
Problem 3A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane
Answer:
5.1 rad/s
Explanation:
Mechanical energy of the system is conserved since no external work is done on the sphere.
[tex]mgh = mv^2/2 + I\omega^2/2[/tex]
Substituting v = ωr and I = 2 m r^2/5, we get,
=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]
=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]
=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]
=> [tex]gh = 7\omega^2 r^2/10[/tex]
=> [tex]\omega r = (10gh/7)^{1/2}[/tex]
=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s
A student writes down several steps of scientific method. Put the steps in the best order
Answer:
Make a hypothesis, conduct an experiment, Analyze the experimental data..
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the vehicle moves for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the ride-sharing car in motion (in s)?
(b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.)
Answer:
Explanation:
Time taken to accelerate to 28 m /s
= 28 / 2 = 14 s
a ) Total length of time in motion
= 14 + 41 + 5
= 60 s .
b )
Distance covered while accelerating
s = ut + 1/2 at²
= 0 + .5 x 2 x 14²
= 196 m .
Distance covered while moving in uniform motion
= 28 x 41
= 1148 m
distance covered while decelerating
v = u - at
0 = 28 - a x 5
a = 5.6 m / s²
v² = u² - 2 a s
0 = 28² - 2 x 5.6 x s
s = 28² / 2 x 5.6
= 70 m .
Total distance covered
= 196 + 1148 + 70
= 1414 m
total time taken = 60 s
average velocity
= 1414 / 60
= 23.56 m /s .
determine the smallest mass of lead that when tied using a string to a wooden boat on a pond will be enough to sink the toy boat. assuming specific gravity of wood is 0.5 and density of water is 1000kg per cubic metre?
Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentum (in kg · m2/s) relative to the point (0, 5.0) meters?
Answer:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
Explanation:
In order to calculate the angular momentum of the particle you use the following formula:
[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex] (1)
r is the position vector respect to the point (0 , 5.0), that is:
r = 0m i + 5.0m j (2)
p is the linear momentum vector and it is given by:
[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex] (3)
the direction of p comes from the fat that the particle is moving along the i + j direction.
Then, you use the results of (2) and (3) in the equation (1) and solve for L:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
The angular momentum is -30 kgm^2/s ^k
still really need help with these three questions!!
Explanation:
2. No, not always. Normal force is equal to force of gravity only when there's no acceleration in the vertical direction.
For example, when you stand in an elevator that's not moving, or moving at constant speed, then the normal force equals your weight. But when the elevator accelerates upward, the normal force increases (making you feel heavier). And when the elevator slows down, the normal force decreases (making you feel lighter).
3. Yes, it is possible for an object to be moving eastward and experience a net force westward. An example is a car applying the brakes.
4. Friction force allows you to walk. When you push against the floor, the floor's friction pushes back, as Newton's third law says.
If you try to walk on a slippery surface like ice, you won't be able to push against the ice, and the ice won't push back.
Two plates with area 7.00×10−3 m27.00×10−3 m2 are separated by a distance of 4.80×10−4 m4.80×10−4 m . If a charge of 5.40×10−8 C5.40×10−8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.
Answer:
The voltage is [tex]V = 418.60 \ Volts[/tex]
Explanation:
From the question we are told that
The area of the both plate is [tex]A = 7.00 *10^{-3} \ m^2[/tex]
The distance between the plate is [tex]d = 4.80*10^{-4}\ m[/tex]
The magnitude of the charge is [tex]q = 5.40 *10^{-8} \ C[/tex]
The capacitance of the capacitor that consist of the two plates is mathematically represented as
[tex]C = \frac{\epsilon _o A}{d}[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value [tex]e = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So
[tex]C = \frac{8.85*10^{-12} * (7* 10^{-3})}{ 4.8*10^{-4}}[/tex]
[tex]C = 1.29 *10^{-10} \ F[/tex]
The potential difference between the plate is mathematically represented as
[tex]V = \frac{ Q}{C }[/tex]
[tex]V = \frac{ 5.4*10^{-8}}{1.29 *10^{-10}}[/tex]
[tex]V = 418.60 \ Volts[/tex]
why is India called peninsula?
Answer:
India is a peninsula.
Explanation:
India is called as Indian Peninsula because it is surrounded by the Indian ocean on the south, the Arabian sea on the west and the Bay of Bengal on the east.
The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one of the following frequencies will NOT resonate in the same pipe
a. 1800 Hz
b. 1000 Hz
c. 1400 Hz
d. 600 Hz
e. 400 Hz
Answer:
e. 400 Hz
Explanation:
In closed organ pipe, only odd harmonics of fundamental note is possible .
The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .
200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc
600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .
Frequency not possible is 400 Hz .
As you get ready for bed, you roll up one of your socks into a tight ball and toss it into the laundry basket across the room. Then, you try to toss the other sock without rolling it up.. What effects whether or not your socks land in the basket?
Answer:
The drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, the size of the sock (weight), and the speed with which the socks is tossed.
Explanation:
The socks, like every other particle or body travelling through air is met by a resistance that impedes its motion. This resistance is due to the air molecules around, that collide with the body as it travels through them. The resistance offered by this force is proportional to the surface area of the body that collides with the air molecule, so, rolling the socks into a ball reduces the effect of air resistance on the socks, compared to the one tossed without rolling. Air resistance is also largely dependent on the relative motion of the body and the air molecules, the density of the fluid (air), and the size of the body (weight).
Therefore, whether the socks lands in the basket or not is affected by the drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, size of the socks (weight), and the speed with which the socks is tossed.
Drag force opposes motion of objects through fluid with its magnitude depending on the velocity of the object in the fluid
The single parameter that effects whether or not the socks lands in the basket is the drag force, [tex]\mathbf{F_D}[/tex] acting on the socks
[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]
The reason that drag force is the parameter that effects the landing point of the socks is as follows:
The parameters that effects whether or not the socks land in the basket or not are;
The distance of the basket away from the thrower = The range, RThe velocity with which the socks are thrown, uThe angle of elevation with which each socks is thrown, θThe amount of drag experienced by each socks, [tex]\mathbf{F_D}[/tex]The parameters, R, u, and θ depends on the thrower, that parameter that effects the whether or not the socks lands in the basket that is independent of the thrower, is the drag, [tex]\mathbf{F_D}[/tex]
Drag is the force opposing (slows) the motion of an object in a fluid.
The drag force, [tex]\mathbf{F_D}[/tex], slowing down motion, is given by the following formula;
[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]
Where;
[tex]v_r[/tex] = The velocity of flow of the fluid, relative to the object
ρ = The density of the fluid
[tex]C_D[/tex] = The drag coefficient
A = The cross sectional area of the fluid
Therefore, the independent parameter that effects whether or not the socks lands in the basket is the drag force on the socks
Learn more about drag force here:
https://brainly.com/question/17074446
A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outstretched hand catches the keys 1.60 s later. (Take upward as the positive direction. Indicate the direction with the sign of your answer.)With what initial velocity were the keys thrown?
Answer:
[tex]v_{i}=10.10 m/s[/tex]
Explanation:
The equation of the position is:
[tex]y=y_{i}+v_{i}t-0.5gt^{2}[/tex]
Where:
v(i) is the initial velocity
The initial position y(i) will be zero and the final position y = 3.60 m.
So, we just need to solve this equation for v(i).
[tex]v_{i}=\frac{y+0.5gt^{2}}{t}[/tex]
[tex]v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}[/tex]
[tex]v_{i}=10.10 m/s[/tex]
Therefore, the initial velocity is 10.10 m/s upwards.
I hope it helps you!
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.
Car A: 500 kg, 10 m/s,
Car B: 2000 kg, 5 m/s,
Car C: 500 kg, 20 m/s,
Car D: 1000 kg, 20 m/s,
Car E: 4000 kg, 5 m/s, and
Car F: 1000 kg, 10 m/s.
(a) Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
(b) Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.
Answer:
a)Car E = Car D > (Car F = Car B = Car C) > Car A
b)Car E = Car D > (Car F = Car B = Car C) > Car A
Explanation:
Car A: mass = 500 kg; speed = 10 m/s
Car B: mass = 2000 kg;speed = 5 m/s
Car C:mass = 500 kg; speed = 20 m/s
Car D: mass = 1000 kg; speed = 20 m/s
Car E:mass = 4000 kg; speed = 5 m/s
Car F: mass = 1000 kg; speed = 10 m/s
Part a) Now we know that momentum of each car is product of mass and velocity , so we will have
CarA:
[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]
Car B:
[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]
Car C:
[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]
Car D:
[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]
Car E:
[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]
Car F:
[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]
So the momentum is given as ,
Car E = Car D > (Car F = Car B = Car C) > Car A
Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car
so the order of impulse from largest to least is given as
Car E = Car D > (Car F = Car B = Car C) > Car A
Chapter 24, Problem 20 GO A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.9 m from his television set. A reporter at the press conference is located 4.1 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Therefore, the distance between politician and TV set is 2536kmExplanation:
Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.
The reporter hears the sound is
4.1 / 343 = 0.01195 s later
The viewer hears the sound from the TV is
2.9 / 343 = 0.00845s
the difference is 0.00845 sec
the question is how far the TV signal can travel in that time.
the distance between politician and TV set is
= 0.00845 * 3*10^8 m
= 2536 km
d = 2536km
Therefore, the distance between politician and TV set is 2536km